SOME PRECISE ESTIMATES OF THE HYPER
ORDER OF SOLUTIONS OF SOME COMPLEX
LINEAR DIFFERENTIAL EQUATIONS
Solutions of Complex Linear
Differential Equations
Benharrat Belaidi
BENHARRAT BELAIDI
Department of Mathematics
Laboratory of Pure and Applied Mathematics
University of Mostaganem
B. P 227 Mostaganem-(Algeria).
EMail: belaidi@univ-mosta.dz
vol. 8, iss. 4, art. 107, 2007
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Received:
05 March, 2007
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Accepted:
30 November, 2007
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Communicated by:
D. Stefanescu
2000 AMS Sub. Class.:
34M10, 30D35.
Key words:
Linear differential equations, Entire solutions, Hyper order.
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Abstract:
Let ρ (f ) and ρ2 (f ) denote respectively the order and the hyper order of an entire
function f. In this paper, we obtain some precise estimates of the hyper order of
solutions of the following higher order linear differential equations
f (k) +
k−1
X
Aj (z) ePj (z) f (j) = 0
j=0
and
f
(k)
+
k−1
X
Solutions of Complex Linear
Pj (z)
Aj (z) e
+ Bj (z) f (j) = 0
j=0
where k ≥ 2, Pj (z) (j = 0, . . . , k − 1) are nonconstant
polynomials
such that deg Pj = n (j = 0, . . . , k − 1) and Aj (z) (6≡ 0) , Bj (z) (6≡ 0)
(j = 0, . . . , k − 1) are entire functions with ρ (Aj ) < n, ρ (Bj ) < n
(j = 0, . . . , k − 1). Under some conditions, we prove that every solution f (z) 6≡ 0
of the above equations is of infinite order and ρ2 (f ) = n.
Differential Equations
Benharrat Belaidi
vol. 8, iss. 4, art. 107, 2007
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Contents
Acknowledgements:
The author would like to thank the referee for his/her helpful remarks and suggestions
to improve the paper.
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Contents
1
Introduction and Statement of Results
4
2
Lemmas Required to Prove Theorem 1.2 and Theorem 1.3
8
3
Proof of Theorem 1.2
11
Solutions of Complex Linear
4
Proof of Theorem 1.3
15
Differential Equations
Benharrat Belaidi
5
Lemmas Required to Prove Theorem 1.4
18
6
Proof of Theorem 1.4
22
vol. 8, iss. 4, art. 107, 2007
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1.
Introduction and Statement of Results
Throughout this paper, we assume that the reader is familiar with the fundamental
results and the standard notations of Nevanlinna’s value distribution theory and with
basic Wiman-Valiron theory as well (see [6], [7], [9], [10]). Let f be a meromorphic
function, one defines
Z 2π
1
m (r, f ) =
log+ f reit dt,
2π 0
Z r
(n (t, f ) − n (0, f ))
N (r, f ) =
dt + n (0, f ) log r,
t
0
and T (r, f ) = m (r, f ) + N (r, f ) (r > 0) is the Nevanlinna characteristic function
of f, where log+ x = max (0, log x) for x ≥ 0 and n (t, f ) is the number of poles of
f (z) lying in |z| ≤ t, counted according to their multiplicity. In addition, we will
T (r, f )
use ρ (f ) = lim loglog
to denote the order of growth of a meromorphic function
r
r→+∞
f (z) . See [6, 9] for notations and definitions.
To express the rate of growth of entire solutions of infinite order, we recall the
following concept.
Definition 1.1 (see [3, 11]). Let f be an entire function. Then the hyper order
ρ2 (f ) of f (z) is defined by
(1.1)
log logT (r, f )
log log log M (r, f )
ρ2 (f ) = lim
= lim
,
r→+∞
r→+∞
log r
log r
where M (r, f ) = max|z|=r |f (z)| .
Several authors have studied the second order linear differential equation
(1.2)
00
0
f + A1 (z) eP1 (z) f + A0 (z) eP0 (z) f = 0,
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Differential Equations
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vol. 8, iss. 4, art. 107, 2007
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where P1 (z) , P0 (z) are nonconstant polynomials, A1 (z) , A0 (z) (6≡ 0) are entire
functions such that ρ (A1 ) < deg P1 (z) , ρ (A0 ) < deg P0 (z). Gundersen showed in
[4, p. 419] that if deg P1 (z) 6= deg P0 (z) , then every nonconstant solution of (1.2)
is of infinite order. If deg P1 (z) = deg P0 (z) , then (1.2) may have nonconstant
00
0
solutions of finite order. For instance f (z) = ez + 1 satisfies f + ez f − ez f = 0.
In [3], Kwon has investigated the order and the hyper order of solutions of equation (1.2) in the case when deg P1 (z) = deg P0 (z) and has obtained the following
result.
Theorem A ([3]). Let P1 (z) and P0 (z) be nonconstant polynomials, such that
(1.3)
P1 (z) = an z n + an−1 z n−1 + · · · + a1 z + a0 ,
(1.4)
P0 (z) = bn z n + bn−1 z n−1 + · · · + b1 z + b0 ,
where ai , bi (i = 0, 1, . . . , n) are complex numbers, an 6= 0, bn 6= 0. Let A1 (z) and
A0 (z) (6≡ 0) be entire functions with ρ (Aj ) < n (j = 0, 1) . Then the following four
statements hold:
(i) If either arg an 6= arg bn or an = cbn (0 < c < 1) , then every nonconstant
solution f of (1.2) has infinite order with ρ2 (f ) ≥ n.
(ii) Let an = bn and deg (P1 − P0 ) = m ≥ 1, and let the orders of A1 (z) and
A0 (z) be less than m. Then every nonconstant solution f of (1.2) has infinite
order with ρ2 (f ) ≥ m.
(iii) Let an = cbn with c > 1 and deg (P1 − cP0 ) = m ≥ 1. Suppose that
ρ (A1 ) < m and A0 (z) is an entire function with 0 < ρ (A0 ) < 1/2. Then
every nonconstant solution f of (1.2) has infinite order with ρ2 (f ) ≥ ρ (A0 ).
(iv) Let an = cbn with c ≥ 1 and let P1 (z) − cP0 (z) be a constant. Suppose that
ρ (A1 ) < ρ (A0 ) < 1/2. Then every nonconstant solution f of (1.2) has infinite
order with ρ2 (f ) ≥ ρ (A0 ).
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Benharrat Belaidi
vol. 8, iss. 4, art. 107, 2007
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In [1], Chen improved the results of Theorem A(i), Theorem A(iii) for the linear
differential equation (1.2) as follows:
P
P
Theorem B ([1]). Let P1 (z) = ni=0 ai z i and P0 (z) = ni=0 bi z i be nonconstant
polynomials where ai , bi (i = 0, 1, . . . , n) are complex numbers, an 6= 0, bn 6= 0,
and let A1 (z) , A0 (z) (6≡ 0) be entire functions. Suppose that either (i) or (ii) below,
holds:
(i) arg an 6= arg bn or an = cbn (0 < c < 1) , ρ (Aj ) < n (j = 0, 1) ;
(ii) an = cbn (c > 1) and deg (P1 − cP0 ) = m ≥ 1, ρ (Aj ) < m (j = 0, 1) .
Then every solution f (z) 6≡ 0 of (1.2) satisfies ρ2 (f ) = n.
Recently, Chen and Shon obtained the following result:
Differential Equations
Benharrat Belaidi
vol. 8, iss. 4, art. 107, 2007
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Theorem C ([2]). Let h0 6≡ 0, h1 , . . . , hk−1 be entire functions with ρ (hj ) < 1
(j = 0, . . . , k − 1). Let a0 6= 0, a1 , . . . , ak−1 be complex numbers such that for
j = 1, . . . , k − 1,
(i) aj = 0, or
(ii) arg aj = arg a0 and aj = cj a0 (0 < cj < 1) , or
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(iii) arg aj 6= arg a0 .
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Then every solution f (z) 6≡ 0 of the linear differential equation
(1.5)
Solutions of Complex Linear
0
f (k) + hk−1 (z) eak−1 z f (k−1) + · · · + h1 (z) ea1 z f + h0 (z) ea0 z f = 0
satisfies ρ (f ) = ∞ and ρ2 (f ) = 1.
The main purpose of this paper is to extend and improve the results of Theorem
B and Theorem C to some higher order linear differential equations. In fact we will
prove the following results.
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Theorem 1.2. Let Pj (z) =
i=0 ai,j z (j = 0, . . . , k − 1) be nonconstant polynomials where a0,j , . . . , an,j (j = 0, 1, . . . , k − 1) are complex numbers such that
an,j an,0 6= 0 (j = 1, . . . , k − 1), and let Aj (z) (6≡ 0) (j = 0, . . . , k − 1) be entire functions. Suppose that arg an,j 6= arg an,0 or an,j = cj an,0 (0 < cj < 1)
(j = 1, . . . , k − 1) and ρ (Aj ) < n (j = 0, . . . , k − 1) . Then every solution f (z) 6≡
0 of the equation
0
(1.6) f (k) + Ak−1 (z) ePk−1 (z) f (k−1) + · · · + A1 (z) eP1 (z) f + A0 (z) eP0 (z) f = 0,
where k ≥ 2, is of infinite order and ρ2 (f ) = n.
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Theorem 1.3. Let Pj (z) =
i=0 ai,j z (j = 0, . . . , k − 1) be nonconstant polynomials where a0,j , . . . , an,j (j = 0, 1, . . . , k − 1) are complex numbers such that
an,j an,0 6= 0 (j = 1, . . . , k−1), and let Aj (z) (6≡ 0) , Bj (z) (6≡ 0) (j = 0, . . . , k − 1)
be entire functions. Suppose that arg an,j 6= arg an,0 or an,j = cj an,0 (0 < cj < 1)
(j = 1, . . . , k − 1) and ρ (Aj ) < n, ρ (Bj ) < n (j = 0, . . . , k − 1) . Then every
solution f (z) 6≡ 0 of the differential equation
(1.7) f (k) + Ak−1 (z) ePk−1 (z) + Bk−1 (z) f (k−1) + · · ·
0
+ A1 (z) eP1 (z) + B1 (z) f + A0 (z) eP0 (z) + B0 (z) f = 0,
where k ≥ 2, is of infinite order and ρ2 (f ) = n.
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Theorem 1.4. Let Pj (z) =
i=0 ai,j z (j = 0, . . . , k − 1) be nonconstant polynomials where a0,j , . . . , an,j (j = 0, 1, . . . , k − 1) are complex numbers such that
an,j an,0 6= 0 (j = 1, . . . , k − 1), and let Aj (z) (6≡ 0) (j = 0, . . . , k − 1) be entire functions. Suppose that an,j = can,0 (c > 1) and deg (Pj − cP0 ) = m ≥ 1
(j = 1, . . . , k − 1), ρ (Aj ) < m (j = 0, . . . , k − 1) . Then every solution f (z) 6≡ 0
of the equation (1.6) is of infinite order and ρ2 (f ) = n.
Solutions of Complex Linear
Differential Equations
Benharrat Belaidi
vol. 8, iss. 4, art. 107, 2007
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2.
Lemmas Required to Prove Theorem 1.2 and Theorem 1.3
We need the following lemmas in the proofs of Theorem 1.2 and Theorem 1.3.
Lemma 2.1 ([5]). Let f (z) be a transcendental meromorphic function, and let α >
1 and ε > 0 be given constants. Then the following two statements hold:
(i) There exists a constant A > 0 and a set E1 ⊂ [0, ∞) having finite linear
measure such that for all z satisfying |z| = r ∈
/ E1 , we have
(j)
f (z) j
ε
(j ∈ N) .
(2.1)
f (z) ≤ A [T (αr, f ) r log T (αr, f )]
(ii) There exists a constant B > 0 and a set E2 ⊂ [0, 2π) that has linear measure
zero, such that if θ ∈ [0, 2π)\E2 , then there is a constant R1 = R1 (θ) > 1
such that for all z satisfying arg z = θ and |z| = r ≥ R1 , we have
(j)
j
f (z) T (αr, f )
α
(log r) log T (αr, f )
(j ∈ N) .
(2.2)
f (z) ≤ B
r
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vol. 8, iss. 4, art. 107, 2007
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n
Lemma 2.2 ([2]). Let P (z) = an z + · · · + a0 , (an = α + iβ 6= 0) be a polynomial
with degree n ≥ 1 and A (z) (6≡ 0) be an entire function with σ (A) < n. Set
f (z) = A (z) eP (z) , z = reiθ , δ (P, θ) = α cos nθ − β sin nθ. Then for any given
ε > 0, there exists a set E3 ⊂ [0, 2π) that has linear measure zero, such that for
any θ ∈ [0, 2π) \ (E3 ∪ E4 ) , where E4 = {θ ∈ [0, 2π) : δ (P, θ) = 0} is a finite set,
there is R2 > 0 such that for |z| = r > R2 , the following statements hold:
(i) if δ (P, θ) > 0, then
(2.3)
exp {(1 − ε) δ (P, θ) rn } ≤ |f (z)| ≤ exp {(1 + ε) δ (P, θ) rn } ,
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(ii) if δ (P, θ) < 0, then
exp {(1 + ε) δ (P, θ) rn } ≤ |f (z)| ≤ exp {(1 − ε) δ (P, θ) rn } .
(2.4)
Lemma 2.3 ([2]). Let A0 (z) , . . . , Ak−1 (z) be entire functions of finite order. If f is
a solution of the equation
(2.5)
0
f (k) + Ak−1 (z) f (k−1) + · · · + A1 (z) f + A0 (z) f = 0,
Solutions of Complex Linear
Differential Equations
then
Benharrat Belaidi
ρ2 (f ) ≤ max {ρ (A0 ) , . . . , ρ (Ak−1 )} .
vol. 8, iss. 4, art. 107, 2007
Lemma 2.4. Let P (z) = bm z m + bm−1 z m−1 + · · · + b1 z + b0 with bm 6= 0 be a
polynomial. Then for every ε > 0, there exists R3 > 0 such that for all |z| = r > R3
the inequalities
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(2.6)
m
(1 − ε) |bm | r ≤ |P (z)| ≤ (1 + ε) |bm | r
m
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hold.
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Proof. Clearly,
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am−1 1
a0 1 |P (z)| = |am | |z| 1 +
+ ··· +
.
am z
am z m m
Denote
am−1 1
a0 1
Rm (z) =
+ ··· +
.
am z
am z m
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Obviously, |Rm (z)| < ε, if |z| > R3 for some ε > 0. This means that
(1 − ε) |am | rm ≤ (1 − |Rm (z)|) |am | rm
≤ |1 + Rm (z)| |am | rm
= |P (z)|
≤ (1 + |Rm (z)|) |am | rm
≤ (1 + ε) |am | rm .
Solutions of Complex Linear
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vol. 8, iss. 4, art. 107, 2007
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3.
Proof of Theorem 1.2
Assume f (z) 6≡ 0 is a transcendental entire solution of (1.6) . By Lemma 2.1 (ii),
there exists a set E2 ⊂ [0, 2π) with linear measure zero, such that if θ ∈ [0, 2π) \E2 ,
there is a constant R1 = R1 (θ) > 1 such that for all z satisfying arg z = θ and
|z| ≥ R1 , we have
(j)
f (z) k+1
(3.1)
(j = 1, . . . , k) , (B > 0) .
f (z) ≤ B [T (2r, f )]
Let P0 (z) = an,0 z n +· · ·+a0,0 (an,0 = α+iβ 6= 0), δ (P0 , θ) = α cos nθ−β sin nθ.
Suppose first that arg an,j 6= arg an,0 (j = 1, . . . , k − 1) . By Lemma 2.2, for any
given ε (0 < ε < 1) , there is a set E3 that has linear measure, and a ray arg z =
θ ∈ [0, 2π) \ (E3 ∪ E4 ), where E4 = {θ ∈ [0, 2π): δ (P0 , θ) = 0 or δ (Pj , θ) =
0 (j = 1, . . . , k − 1)} (E4 is a finite set), such that δ (P0 , θ) > 0, δ (Pj , θ) < 0
(j = 1, . . . , k − 1) and for sufficiently large |z| = r, we have
A0 (z) eP0 (z) ≥ exp {(1 − ε) δ (P0 , θ) rn }
(3.2)
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vol. 8, iss. 4, art. 107, 2007
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and
(3.3)
Solutions of Complex Linear
Aj (z) ePj (z) ≤ exp {(1 − ε) δ (Pj , θ) rn } < 1
(j = 1, . . . , k − 1) .
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It follows from (1.6) that
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(3.4)
(k)
(k−1)
f
(z)
f
(z)
A0 (z) eP0 (z) ≤ + Ak−1 (z) ePk−1 (z) f (z) f (z) 0
P1 (z) f (z) + · · · + A1 (z) e
f (z) .
Now, take a ray θ ∈ [0, 2π) \ (E2 ∪ E3 ∪ E4 ) . Hence by (3.1) − (3.3) and (3.4) , we
get for sufficiently large |z| = r
(3.5)
exp {(1 − ε) δ (P0 , θ) rn }
≤
1+
k−1
X
!
exp {(1 − ε) δ (Pj , θ) rn } B [T (2r, f )]k+1
j=1
k+1
≤ kB [T (2r, f )]
Solutions of Complex Linear
.
By 0 < ε < 1 and (3.5) we get that ρ (f ) = +∞ and
(3.6)
log logT (r, f )
≥ n.
r→+∞
log r
Benharrat Belaidi
vol. 8, iss. 4, art. 107, 2007
ρ2 (f ) = lim
By Lemma 2.3, we have ρ2 (f ) = n.
Suppose now an,j = cj an,0 (0 < cj < 1) (j = 1, . . . , k − 1) . Then δ (Pj , θ) =
cj δ (P0 , θ) (j = 1, . . . , k − 1) . Put c = max{cj : j = 1, . . . , k − 1}. Then 0 <
1−c
c < 1. Using the same reasoning as above, for any given ε (0 < 2ε < 1+c
) there
exists a ray arg z = θ ∈ [0, 2π) \ (E5 ∪ E6 ), where E5 and E6 are defined as in
Lemma 2.2, E5 ∪ E6 is of linear measure zero, satisfying δ (Pj , θ) = cj δ (P0 , θ) > 0
(j = 1, . . . , k − 1) and for sufficiently large |z| = r, we have
A0 (z) eP0 (z) ≥ exp {(1 − ε) δ (P0 , θ) rn }
(3.7)
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and
(3.8)
Differential Equations
Aj (z) ePj (z) ≤ exp {(1 + ε) δ (Pj , θ) rn }
≤ exp {(1 + ε) cδ (P0 , θ) rn } (j = 1, . . . , k − 1) .
Now, take a ray θ ∈ [0, 2π) \ (E2 ∪ E5 ∪ E6 ) . By substituting (3.1) , (3.7) and (3.8)
into (3.4), we get for sufficiently large |z| = r,
(3.9)
exp {(1 − ε) δ (P0 , θ) rn }
≤ (1 + (k − 1) exp {(1 + ε) cδ (P0 , θ) rn }) B [T (2r, f )]k+1
≤ kB exp {(1 + ε) cδ (P0 , θ) rn } [T (2r, f )]k+1 .
Solutions of Complex Linear
By 0 < 2ε <
(3.10)
1−c
1+c
and (3.9) we have
(1 − c)
n
δ (P0 , θ) r
≤ kB [T (2r, f )]k+1 .
exp
2
Thus, (3.10) implies ρ (f ) = +∞ and
(3.11)
log logT (r, f )
≥ n.
r→+∞
log r
ρ2 (f ) = lim
By Lemma 2.3, we have ρ2 (f ) = n.
Now we prove that equation (1.6) cannot have a nonzero polynomial solution.
Suppose first that arg an,j 6= arg an,0 (j = 1, . . . , k − 1) . Assume f (z) 6≡ 0 is a
polynomial solution of (1.6) . By Lemma 2.2, for any given ε (0 < ε < 1) there
exists a ray arg z = θ ∈ [0, 2π) \ (E3 ∪ E4 ) satisfying δ (P0 , θ) > 0, δ (Pj , θ) < 0
(j = 1, . . . , k − 1) and for sufficiently large |z| = r, inequalities (3.2) , (3.3) hold.
By (1.6) we can write
(3.12) A0 (z) eP0 (z) f = − f (k) + Ak−1 (z) ePk−1 (z) f (k−1) + · · · + A1 (z) eP1 (z) f 0 .
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vol. 8, iss. 4, art. 107, 2007
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By using (3.2) , (3.3) , (3.12) and Lemma 2.4 we obtain for sufficiently large |z| = r
(3.13)
(1 − ε) |bm | rm exp {(1 − ε) δ (P0 , θ) rn }
≤ A0 (z) eP0 (z) f (k)
Pk−1 (z) (k−1)
P1 (z) 0 = f + Ak−1 (z) e
f
+ · · · + A1 (z) e
f
≤ k (1 + ε) m |bm | rm−1 ,
Solutions of Complex Linear
where f (z) = bm z m + bm−1 z m−1 + · · · + b1 z + b0 with bm 6= 0. From (3.13) we get
for sufficiently large |z| = r
(3.14)
exp {(1 − ε) δ (P0 , θ) rn } ≤ k
1+ε 1
m .
1−ε r
This is absurd since 0 < ε < 1. By using similar reasoning as above we can prove
that if an,j = cj an,0 (0 < cj < 1) , then equation (1.6) cannot have nonzero polynomial solution. Hence every solution f (z) 6≡ 0 of (1.6) is of infinite order and
ρ2 (f ) = n.
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vol. 8, iss. 4, art. 107, 2007
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4.
Proof of Theorem 1.3
Assume f (z) 6≡ 0 is a solution of (1.7) . By using similar reasoning as in the proof of
Theorem 1.2, it follows that f (z) must be a transcendental entire solution. Suppose
first that arg an,j 6= arg an,0 (j = 1, . . . , k − 1). By Lemma 2.2, for any given ε
(0 < 2ε < min {1, n − α}), where α = max{ρ (Bj ) : j = 0, . . . , k − 1}, there
exists a ray arg z = θ such that θ ∈ [0, 2π) \(E3 ∪E4 ), where E3 and E4 are defined
as in Lemma 2.2, E3 ∪ E4 is of linear measure zero, and δ (P0 , θ) > 0, δ (Pj , θ) < 0
(j = 1, . . . , k − 1) and for sufficiently large |z| = r, we have
A0 (z) eP0 (z) + B0 (z) ≥ (1 − o (1)) exp {(1 − ε) δ (P0 , θ) rn }
(4.1)
and
(4.2)
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Benharrat Belaidi
vol. 8, iss. 4, art. 107, 2007
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Aj (z) ePj (z) + Bj (z) ≤ exp {(1 − ε) δ (Pj , θ) rn } + exp rρ(Bj )+ 2ε
ρ(Bj )+ε
≤ exp r
n α+ε o
≤ exp r
(j = 1, . . . , k − 1) .
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It follows from (1.7) that
(4.3) A0 (z) eP0 (z) + B0 (z)
(k)
f (z) f (k−1) (z) Pk−1 (z)
+ ···
≤
+ Ak−1 (z) e
+ Bk−1 (z) f (z) f (z) f 0 (z) P
(z)
1
.
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Now, take a ray θ ∈ [0, 2π) \ (E2 ∪ E3 ∪ E4 ) . Hence by (3.1) and (4.1) − (4.3) , we
get for sufficiently large |z| = r
(4.4)
(1 − o (1)) exp {(1 − ε) δ (P0 , θ) rn }
n α+ε o
≤ 1 + (k − 1) exp r
B [T (2r, f )]k+1
n α+ε o
≤ kB exp r
[T (2r, f )]k+1 .
Solutions of Complex Linear
Differential Equations
Thus, 0 < 2ε < min {1, n − α} implies ρ (f ) = +∞ and
(4.5)
log logT (r, f )
≥ n.
ρ2 (f ) = lim
r→+∞
log r
By Lemma 2.3, we have ρ2 (f ) = n.
Suppose now an,j = cj an,0 (0 < cj < 1) (j = 1, . . . , k − 1) . Then δ (Pj , θ) =
cj δ (P0 , θ) (j = 1, . . . , k − 1) . Put c = max{cj : j = 1, . . . , k−1}. Then
0 < c < 1.
1−c
Using the same reasoning as above, for any given ε 0 < 2ε < 1+c there exists a
ray arg z = θ ∈ [0, 2π) \(E5 ∪ E6 ), E5 ∪ E6 is of linear measure zero, satisfying
δ (Pj , θ) = cj δ (P0 , θ) > 0 (j = 1, . . . , k − 1) and for sufficiently large |z| = r, we
have
A0 (z) eP0 (z) + B0 (z) ≥ (1 − o (1)) exp {(1 − ε) δ (P0 , θ) rn }
(4.6)
and
(4.7)
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Aj (z) ePj (z) + Bj (z)
≤ (1 + o (1)) exp {(1 + ε) cδ (P0 , θ) rn } (j = 1, . . . , k − 1) .
Now, take a ray θ ∈ [0, 2π) \ (E2 ∪ E5 ∪ E6 ) . By substituting (3.1) , (4.6) and (4.7)
into (4.3), we get for sufficiently large |z| = r,
(4.8)
(1 − o (1)) exp {(1 − ε) δ (P0 , θ) rn }
≤ (1 + (k − 1) (1 + o (1)) exp {(1 + ε) cδ (P0 , θ) rn }) B [T (2r, f )]k+1
≤ kB (1 + o (1)) exp {(1 + ε) cδ (P0 , θ) rn } [T (2r, f )]k+1 .
By 0 < 2ε <
(4.9)
1−c
1+c
and (4.8) we have
(1 − c)
n
exp
δ (P0 , θ) r
≤ kBd [T (2r, f )]k+1 ,
2
Solutions of Complex Linear
Differential Equations
Benharrat Belaidi
vol. 8, iss. 4, art. 107, 2007
where d > 0 is some constant. Thus, (4.9) implies ρ (f ) = +∞ and
(4.10)
log logT (r, f )
≥ n.
ρ2 (f ) = lim
r→+∞
log r
By Lemma 2.3, we have ρ2 (f ) = n.
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5.
Lemmas Required to Prove Theorem 1.4
We need the following lemmas in the proof of Theorem 1.4.
P
Lemma 5.1 ([8, pp. 253-255]). Let P0 (z) = ni=0 bi z i , where n is a positive integer
and bn = αn eiθn , αn > 0, θn ∈ [0, 2π) . For any given ε (0 < ε < π/4n), we
introduce 2n closed angles
θn
π
θn
π
+ ε ≤ θ ≤ − + (2j + 1)
−ε
Sj : − + (2j − 1)
n
2n
n
2n
(j = 0, 1, . . . , 2n − 1) .
(5.1)
Solutions of Complex Linear
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Benharrat Belaidi
vol. 8, iss. 4, art. 107, 2007
Then there exists a positive number R4 = R4 (ε) such that for |z| = r > R4 ,
Re P0 (z) > αn rn (1 − ε) sin (nε) ,
(5.2)
Contents
if z = reiθ ∈ Sj , when j is even; while
Re P0 (z) < −αn rn (1 − ε) sin (nε) ,
(5.3)
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iθ
if z = re ∈ Sj , when j is odd.
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− θnn
π
2n
Now for any given θ ∈ [0, 2π) , if θ 6=
+ (2j − 1) (j = 0, 1, . . . , 2n − 1),
then we take ε sufficiently small, and there is some Sj (j = 0, 1, . . . , 2n − 1) such
that z = reiθ ∈ Sj .
Lemma 5.2 ([1]). Let f (z) be an entire function of order ρ (f ) = α < +∞. Then
for any given ε > 0, there exists a set E7 ⊂ [1, +∞) that has finite linear measure
and finite logarithmic measure, such that for all z satisfying |z| = r ∈
/ [0, 1] ∪ E7 ,
we have
(5.4)
exp −rα+ε ≤ |f (z)| ≤ exp rα+ε .
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P
n
Lemma 5.3 ([1]). Let f (z) = ∞
n=0 an z be an entire function of infinite order with
the hyper order ρ2 (f ) = σ, µ (r) be the maximum term, i.e., µ (r) = max{|an | rn ;
n = 0, 1, . . . } and let νf (r) be the central index of f , i.e., νf (r) = max{m, µ (r) =
|am | rm }. Then
(5.5)
log log νf (r)
= σ.
r→+∞
log r
lim
Lemma 5.4 (Wiman-Valiron, [7, 10]). Let f (z) be a transcendental entire function
and let z be a point with |z| = r at which |f (z)| = M (r, f ). Then for all |z| outside
a set E8 of r of finite logarithmic measure, we have
j
f (j) (z)
νf (r)
(5.6)
=
(1 + o (1))
(j is an integer, r ∈
/ E8 ).
f (z)
z
Lemma 5.5 ([1]). Let f (z) be an entire function with ρ (f ) = +∞ and ρ2 (f ) =
α < +∞, leta set E9 ⊂ [1, +∞) have finite logarithmic measure. Then there exists
zp = rp ei θp such that |f (zp )| = M (rp , f ) , θp ∈ [0, 2π) , limp→+∞ θp = θ0 ∈
[0, 2π) , rp ∈
/ E9 , rp → +∞, and for any given ε > 0, for sufficiently large rp , we
have
log νf (rp )
(5.7)
lim
= +∞,
p→+∞
log rp
(5.8)
exp rpα−ε ≤ νf (rp ) ≤ exp rpα+ε .
P
Lemma 5.6. Let Pj (z) = ni=0 ai,j z i (j = 0, . . . , k − 1) be nonconstant polynomials where a0,j , . . . , an,j (j = 0, 1, . . . , k − 1) are complex numbers such that
an,j an,0 6= 0 (j = 1, . . . , k − 1), let Aj (z) (6≡ 0) (j = 0, . . . , k − 1) be entire functions. Suppose that an,j = can,0 (c > 1) and deg (Pj − cP0 ) = m ≥ 1 (j = 1, . . . , k − 1),
ρ (Aj ) < m (j = 0, . . . , k − 1) . Then every solution f (z) 6≡ 0 of the equation (1.6)
is of infinite order and ρ2 (f ) ≥ m.
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Proof. Assume f (z) 6≡ 0 is a solution of (1.6) . By using similar reasoning as in the
proof of Theorem 1.2, it follows that f (z) must be a transcendental entire solution.
From (1.6) , we have
(5.9) A0 (z) e(1−c)P0 (z) (k−1)
f
−cP (z) f (k) (z) (z)
≤ e 0 + Ak−1 (z) ePk−1 (z)−cP0 (z) f (z) f (z) f 0 (z) P
(z)−cP
(z)
1
0
+ · · · + A1 (z) e
f (z) .
By Lemma 2.1 (i), there exists a constant A > 0 and a set E1 ⊂ [0, ∞) having finite
linear measure such that for all z satisfying |z| = r ∈
/ E1 , we have
(j)
f (z) k+1
(5.10)
(j = 1, . . . , k) .
f (z) ≤ Ar [T (2r, f )]
By (5.9) and (5.10) , we have for all z satisfying |z| = r ∈
/ E1
(5.11) A0 (z) e(1−c)P0 (z) ≤ e−cP0 (z) + Ak−1 (z) ePk−1 (z)−cP0 (z) + · · ·
+ A1 (z) eP1 (z)−cP0 (z) Ar [T (2r, f )]k+1 .
Since deg (Pj − cP0 ) = m < deg P0 = n (j = 1, . . . , k − 1) , by Lemma 5.1 (see
also [3, p. 385]), there exists a positive real number b and a curve Γ tending to
infinity such that for all z ∈ Γ with |z| = r, we have
(5.12)
Re P0 (z) = 0,
Re (Pj (z) − cP0 (z)) ≤ −brm
(j = 1, . . . , k − 1) .
Let max {ρ (Aj ) (j = 0, . . . , k − 1)} = β < m. Then by Lemma 5.2, there exists
a set E7 ⊂ [1, +∞) that has finite linear measure, such that for all z satisfying
Solutions of Complex Linear
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|z| = r ∈
/ [0, 1] ∪ E7 , we have
(5.13)
exp −rβ+ε ≤ |Aj (z)| ≤ exp rβ+ε
(j = 0, . . . , k − 1) .
Hence by (5.11) − (5.13) , we get for all z ∈ Γ with |z| = r ∈
/ [0, 1] ∪ E1 ∪ E7
β+ε β+ε (5.14) exp −r
≤ 1 + (k − 1) exp r
exp {−brm } Ar [T (2r, f )]k+1 .
Solutions of Complex Linear
Thus β + ε < m implies ρ (f ) = +∞ and
log logT (r, f )
ρ2 (f ) = lim
≥ m.
r→+∞
log r
Differential Equations
Benharrat Belaidi
vol. 8, iss. 4, art. 107, 2007
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6.
Proof of Theorem 1.4
Assume f (z) 6≡ 0 is a solution of (1.6) . Then by Lemma 5.6 and Lemma 2.3, we
have ρ (f ) = ∞ and m ≤ ρ2 (f ) ≤ n. We show that ρ2 (f ) = n. We assume that
ρ2 (f ) = λ (m ≤ λ < n) , and we prove that ρ2 (f ) = λ fails. By the Wiman-Valiron
theory, there is a set E8 ⊂ [1, +∞) with logarithmic measure lm (E8 ) < +∞ and
we can choose z satisfying |z| = r ∈
/ [0, 1] ∪ E8 and |f (z)| = M (r, f ), such that
(5.6) holds. Set max{ρ (Aj ) (j = 0, . . . , k − 1)} = β < m. By Lemma 5.2, for any
given ε (0 < 3ε < min (m − β, n − λ)), there exists a set E7 ⊂ [1, +∞) that has
finite logarithmic measure, such that for all z satisfying |z| = r ∈
/ [0, 1] ∪ E7 , the
inequalities (5.13) hold and
(6.1)
exp −rm+ε ≤ |exp {Pj (z) − cP0 (z)}|
≤ exp rm+ε
(j = 1, . . . , k − 1) .
By Lemma 5.5, we can choose a point range zp = rp ei θp such that |f (zp )| =
M (rp , f ) , θp ∈ [0, 2π) , limp→+∞ θp = θ0 ∈ [0, 2π) , rp ∈
/ [0, 1] ∪ E7 ∪ E8 , rp →
+∞, and for the above ε > 0, for sufficiently large rp , we have
(6.2)
exp rpλ−ε ≤ νf (rp ) ≤ exp rpλ+ε ,
log νf (rp )
(6.3)
lim
= +∞.
p→+∞
log rp
Pn
i
iθn
Let P0 (z) =
, αn > 0.
i=0 bi z where n is a positive integer and bn = αn e
By Lemma 5.1, for any given ε (0 < 3ε < min(m − β, n − λ, π/4n)), there are 2n
closed angles
θn
π
θn
π
+(2j − 1) +ε ≤ θ ≤ − +(2j + 1) −ε
(j = 0, 1, . . . , 2n − 1) .
n
2n
n
2n
π
For the above θ0 and 0 < 3ε < min m − β, n − λ, 4n
, there are three cases:
Sj : −
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Benharrat Belaidi
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1. rp ei θ0 ∈ Sj where j is odd;
2. rp ei θ0 ∈ Sj where j is even;
π
3. θ0 = − θnn + (2j − 1) 2n
for some j = 0, 1, . . . , 2n − 1.
Now we have three cases to prove Theorem 1.4.
Case (1): rp ei θ0 ∈ Sj where j is odd. Since limp→+∞ θp = θ0 , there is a N > 0
such that rp ei θp ∈ Sj when p > N. By Lemma 5.1, we have
(6.4) Re P0 rp ei θp < −δrpn (δ > 0) , i.e.,
Re −P0 rp ei θp > δrpn .
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By (1.6), we have
− e−P0 (z)
(k−1)
f
f
= Ak−1 (z) ePk−1 (z)−P0 (z)
f
f
+ A1 (z) e
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+ ···
P1 (z)−P0 (z) f
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0
f
+ A0 (z) .
Substituting (5.6) into (6.6) , we get for zp = rp ei θp
(6.7)
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= Re {(c − 1) P0 + (Pj − cP0 )}
< rpm+ε − (c − 1) δrpn
(j = 1, . . . , k − 1) .
(6.6)
Differential Equations
vol. 8, iss. 4, art. 107, 2007
From (6.1) and (6.4) , we obtain for sufficiently large p,
(6.5)
Re Pj rp ei θp − P0 rp ei θp
(k)
Solutions of Complex Linear
− νfk (rp ) (1 + o (1)) exp {−P0 (zp )}
= Ak−1 (zp ) exp {Pk−1 (zp ) − P0 (zp )} zp νfk−1 (rp ) (1 + o (1)) + · · ·
+ A1 (zp ) exp {P1 (zp ) − P0 (zp )} zpk−1 νf (rp ) (1 + o (1)) + zpk A0 (zp ) .
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Thus we have, from (6.2) and (6.4)
(6.8) −νfk (rp ) (1 + o (1)) exp {−P0 (zp )}
1
≥ exp δrpn exp krpλ−ε > exp δrpn .
2
And by (5.13) , (6.5) and (6.2) , we have
(6.9) Ak−1 (zp ) exp {Pk−1 (zp ) − P0 (zp )} zp νfk−1 (rp ) (1 + o (1)) + · · ·
+A1 (zp ) exp {P1 (zp ) − P0 (zp )} zpk−1 νf (rp ) (1 + o (1)) + zpk A0 (zp )
≤ 2 (k − 1) rpk−1 exp rpβ+ε exp rpm+ε − (c − 1) δrpn
× exp (k − 1) rpλ+ε + rpk exp rpβ+ε
≤ exp (k − 1) rpλ+2ε .
From (6.7) we see that (6.8) contradicts (6.9) .
Case (2): rp ei θ0 ∈ Sj where j is even. Since limp→+∞ θp = θ0 , there is a N > 0
such that rp ei θp ∈ Sj when p > N. By Lemma 5.1, we have
(6.10)
Re P0 rp ei θp > δrpn , Re −cP0 rp ei θp < −cδrpn ,
(6.11) Re (1 − c) P0
rp ei θp < (1 − c) δrpn , Re Pj rp ei θp − cP0 rp ei θp
<
−cδrpn
(j = 2, . . . , k − 1) .
Benharrat Belaidi
vol. 8, iss. 4, art. 107, 2007
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0
− A1 (z) eP1 (z)−cP0 (z)
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By (1.6), we have
(6.12)
Solutions of Complex Linear
(k)
(k−1)
f
f
f
= e−cP0 (z)
+ Ak−1 (z) ePk−1 (z)−cP0 (z)
f
f
f
00
f
+ · · · + A2 (z) eP2 (z)−cP0 (z)
+ A0 (z) e(1−c)P0 (z) .
f
Substituting (5.6) into (6.12) we get for zp = rp ei θp
(6.13)
− A1 (zp ) exp {P1 (zp ) − cP0 (zp )} zpk−1 νf (rp ) (1 + o (1))
= νfk (rp ) (1 + o (1)) exp {−cP0 (zp )} + νfk−1 (rp ) (1 + o (1))
× zp Ak−1 (zp ) exp {Pk−1 (zp ) − cP0 (zp )} + · · · + zpk−2 νf2 (rp ) (1 + o (1))
× A2 (zp ) exp {P2 (zp ) − cP0 (zp )} + zpk A0 (zp ) exp {(1 − c) P0 (zp )} .
Thus we get, from (5.13) , (6.1) and (6.2)
(6.14) −A1 (zp ) exp {P1 (zp ) − cP0 (zp )} zpk−1 νf (rp ) (1 + o (1))
1
≥ rpk−1 exp −rpβ+ε exp −rpm+ε exp rpλ−ε > exp −rpm+ε .
2
And from (5.13) , (6.2) and (6.10) , (6.11) we have for sufficiently large p
(6.15) νfk (rp ) (1 + o (1)) exp {−cP0 (zp )} + νfk−1 (rp ) (1 + o (1))
× zp Ak−1 (zp ) exp {Pk−1 (zp ) − cP0 (zp )} + · · · + zpk−2 νf2 (rp ) (1 + o (1))
×A2 (zp ) exp {P2 (zp ) − cP0 (zp )} + zpk A0 (zp ) exp {(1 − c) P0 (zp )}
≤ 2 exp krpλ+ε exp −cδrpn + 2 (k − 2) rpk−2 exp (k − 1) rpλ+ε
× exp rpβ+ε exp −cδrpn + rpk exp rpβ+ε exp (1 − c) δrpn
(1 − c) n
< exp
δrp .
2
Thus (6.13) − (6.15) imply a contradiction.
π
Case (3) : θ0 = − θnn +(2j − 1) 2n
for some j = 0, 1, . . . , 2n−1. Since Re P0 rp ei θ0
= 0 when rp is sufficiently large and a straight line arg z = θ0 is an asymptotic line
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of rp ei θp , there is a N > 0 such that when p > N , we have
(6.16)
−1 < Re P0 rp eiθp < 1,
−c < Re Pj rp ei θp < c
(j = 1, . . . , k − 1) .
− cP0 rp ei θp , we again split this into three cases.
By considering Re Pj rp ei θp
Case (i):
(6.17)
Re Pj rp ei θp − cP0 rp ei θp < −drpm
(j = 1, . . . , k − 1)
(d > 0 is a constant) when p is sufficiently large. We have a slightly modified form
of (6.7)
(6.18)
− νfk (rp ) (1 + o (1)) exp {−cP0 (zp )}
= Ak−1 (zp ) exp {Pk−1 (zp ) − cP0 (zp )} zp νfk−1 (rp ) (1 + o (1)) + · · ·
+ A1 (zp ) exp {P1 (zp ) − cP0 (zp )} zpk−1 νf (rp ) (1 + o (1))
+ zpk A0 (zp ) exp {(1 − c) P0 (zp )} .
Thus we get, from (5.13) and (6.16) − (6.18)
1 k
νf (rp ) exp {−c} < −νfk (rp ) (1 + o (1)) exp {−cP0 (zp )}
2
≤ Ak−1 (zp ) exp {Pk−1 (zp ) − cP0 (zp )} zp νfk−1 (rp ) (1 + o (1))
+ · · · + A1 (zp ) exp {P1 (zp ) − cP0 (zp )} zpk−1 νf (rp ) (1 + o (1))
+ zpk A0 (zp ) exp {(1 − c) P0 (zp )}
≤ 2 (k − 1) rpk−1 νfk−1 (rp ) exp rpβ+ε − drpm
+ rpk exp rpβ+ε exp {(c − 1)}
≤ νfk−1 (rp ) exp rpβ+2ε
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i.e.,
1
νf (rp ) < exp {c} exp rpβ+2ε .
2
This is in contradiction with νf (rp ) ≥ exp rpλ−ε .
Case (ii):
Re Pj rp ei θp − cP0 rp ei θp > drpm
(j = 1, . . . , k − 1) ,
Solutions of Complex Linear
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i.e.,
Benharrat Belaidi
(6.19)
Re P0
1
rp ei θp − Pj rp ei θp
c
d
< − rpm
c
(j = 1, . . . , k − 1)
(d > 0 is a constant) when p is sufficiently large. From (6.16) , we obtain for
sufficiently large p,
1
i θp
i θp
(6.20)
Re Ps rp e
− Pj rp e
c
1
= Re Ps rp ei θp − Re Pj rp ei θp
c
< c + 1 (s = 1, . . . , j − 1, j + 1, . . . , k − 1) .
We have a slightly modified form of (6.7)
1
k
(6.21) − νf (rp ) (1 + o (1)) exp − Pj (zp )
c
1
= Ak−1 (zp ) exp Pk−1 (zp ) − Pj (zp ) zp νfk−1 (rp ) (1 + o (1)) + · · ·
c
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1
+ Aj (zp ) exp
1−
Pj (zp ) zpk−j νfj (rp ) (1 + o (1)) + · · ·
c
1
+ A1 (zp ) exp P1 (zp ) − Pj (zp ) zpk−1 νf (rp ) (1 + o (1))
c
1
k
+ zp A0 (zp ) exp P0 (zp ) − Pj (zp ) .
c
Thus we get, from (5.13) , (6.16) and (6.19) − (6.21)
Solutions of Complex Linear
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Benharrat Belaidi
1 k
ν (rp ) exp {−1}
2 f k
1
< −νf (rp ) (1 + o (1)) exp − Pj (zp ) c
1
k−1
≤ Ak−1 (zp ) exp Pk−1 (zp ) − Pj (zp ) zp νf (rp ) (1 + o (1))
c
1
j
+ · · · + Aj (zp ) exp
1−
Pj (zp ) zpk−j νf (rp ) (1 + o (1))
c
1
k−1
+ · · · + A1 (zp ) exp P1 (zp ) − Pj (zp ) zp νf (rp ) (1 + o (1))
c
k
1
+ zp A0 (zp ) exp P0 (zp ) − Pj (zp ) c
β+ε k−1
k−1
≤ 2 (k − 1) νf (rp ) rp exp rp
β+ε d m
k
× exp {c + 1} + rp exp rp
exp − rp
c
β+2ε k−1
≤ νf (rp ) exp rp
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i.e.,
νf (rp ) exp {−1} < 2 exp rpβ+2ε .
This is in contradiction with νf (rp ) ≥ exp rpλ−ε .
Case (iii): When p is sufficiently large,
(j = 1, . . . , k − 1) .
−1 < Re Pj rp ei θp − cP0 rp ei θp < 1
Solutions of Complex Linear
By using the same reasoning as in Case (ii) we get a contradiction. The proof of
Theorem 1.4 is completed.
Differential Equations
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vol. 8, iss. 4, art. 107, 2007
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References
[1] Z.X. CHEN, On the hyper order of solutions of some second order linear differential equations, Acta Mathematica Sinica Engl. Ser., 18(1) (2002), 79–88.
[2] Z.X. CHEN AND K.H. SHON, The growth of solutions of higher order differential equations, Southeast Asian Bull. Math., 27 (2004), 995–1004.
[3] K.H. KWON, Nonexistence of finite order solutions of certain second order
linear differential equations, Kodai Math. J., 19 (1996), 378–387.
[4] G.G. GUNDERSEN, Finite order solutions of second order linear differential
equations, Trans. Amer. Math. Soc., 305 (1988), 415–429.
[5] G.G. GUNDERSEN, Estimates for the logarithmic derivative of a meromorphic
function, plus similar estimates, J. London Math. Soc., (2) 37 (1988), 88–104.
[6] W.K. HAYMAN, Meromorphic Functions, Clarendon Press, Oxford, 1964.
[7] W.K. HAYMAN, The local growth of power series: a survey of the WimanValiron method, Canad. Math. Bull., 17 (1974), 317–358.
[8] A.I. MARKUSHEVICH, Theory of Functions of a Complex Variable, Vol. II,
translated by R.A. Silverman, Prentice-Hall, Englewood Cliffs, New Jersey,
1965.
[9] R. NEVANLINNA, Eindeutige Analytische Funktionen, Zweite Auflage.
Reprint. Die Grundlehren der mathematischen Wissenschaften, Band 46.
Springer-Verlag, Berlin-New York, 1974.
[10] G. VALIRON, Lectures on the General Theory of Integral Functions, translated
by E.F. Collingwood, Chelsea, New York, 1949.
[11] H.X. YI AND C.C. YANG, The Uniqueness Theory of Meromorphic Functions,
Science Press, Beijing, 1995 (in Chinese).
Solutions of Complex Linear
Differential Equations
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